Question

If 2tanA = 3tanB, prove that tan(A-B) = (sin2A)/(5-cos2B)

Answer 1

tan(A-B) = (tanA - tanB)/(1 + tanAtanB)

= (3/2 tanB - tanB)/(1 + 3/2 tanBtanB)      ... given 2tanA = 3tanB

= tanB(3/2  - 1)/(1 + 3/2 tan2B)    

= tanB (1/2)/(sec2B - tan2B + 3/2 tan2B)      .. since 1 + tan2B = sec2B

= tanB (1/2)/(sec2B + 1/2 tan2B)    

= sinB cosB/(2 +  sin2B)

= 2sinB cosB/(4 +  2sin2B) 
= sin2B /(4 +  1 - cos2B)                              .... 2sin2B = 1 - cos2B
= sin2B /(5 - cos2B)